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Integrating factors problem

  1. May 13, 2007 #1

    ssb

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    1. The problem statement, all variables and given/known data

    [tex]y' + 2ty = t^3[/tex]

    2. Relevant equations

    Integrating factors and variation of parameters

    3. The attempt at a solution

    Ive solved for m
    [tex] M = e^{\int 2t\,dx} [/tex]

    [tex] M = e^t^2 [/tex] (this is e^t^2, but doesnt look like it in latex)

    I multiplied both sides by M

    [tex](e^t^2)(y') + (e^t^2)(2ty) = (e^t^2)(t^3)[/tex]

    The part im having trouble with is the next step. Im suppost to find a function so when you take the derivative of that function, it is the same as the left hand side of this problem, or [tex](e^t^2)(y') + (e^t^2)(2ty)[/tex]

    Any suggestions? Is there a simple formula you can use that always works in these cases? It seems like ive tried [tex]\frac{d}{dt}(y)(e^t^2)[/tex] but it wont work.
     
    Last edited: May 13, 2007
  2. jcsd
  3. May 13, 2007 #2
    Just pick the undifferentiated part from each term on the left hand side, ie,

    [itex]\left( e^{t^2} y \right)' = t^3 e^{t^2}[/itex]
     
  4. May 13, 2007 #3

    ssb

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    Thats exactly what ive been doing yet webworks is rejecting my work. Let me provide the rest of my steps just to verify:

    You know what... since im over the deadline webworks gives me the option to show answers... my answers im suppost to be putting in are all jacked up. Looks like the teacher messed up programming it this week for me. Ive been doing it right all along!!! Thanks for the help buddy!
     
  5. May 13, 2007 #4

    ssb

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    What if I had

    [tex]t^3y' - (3/t)yt^3 = t^6 [/tex]


    [tex](?)' = t^6[/tex]


    would it be [tex]yt^3[/tex] ????
     
  6. May 13, 2007 #5
    Well, we can check ... [itex](t^3y)' = 3t^2y + t^3y'[/itex], so other than the minus sign, I think you're good.
     
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